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Đặt : $u = {\ln ^2}x\,\,\,\,\, \Rightarrow \,\,\,\,du = \frac{{2\ln x\,dx}}{x}$
$dv = \frac{{dx}}{{{x^2}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,v = \frac{{ - 1}}{x}$
${J_{\left(
t \right)}} = - \left. {\frac{1}{x}{{\ln }^2}x} \right|_1^t +
\,\,2\int\limits_1^t {\frac{{\ln x}}{{{x^2}}}dx = - \frac{1}{t}{{\ln
}^2}t + 2} \int\limits_1^t {\frac{{\ln x}}{{{x^2}}}dx} $
- Đặt $u = \ln x\,\,\, \Rightarrow \,\,\,du = \frac{{dx}}{x}$
$dv = \frac{{dx}}{{{x^2}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,v = \frac{{ - 1}}{x}$
Suy ra: $\int\limits_1^t {\frac{{\ln x}}{{{x^2}}}dx} = - \left.
{\frac{1}{x}\ln x} \right|_1^t + \int\limits_1^t {\frac{1}{{{x^2}}}dx} $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \left. {\frac{1}{t}\ln t
- \frac{1}{x}} \right|_1^t = - \frac{1}{t}\ln t - \frac{1}{t} + 1$
Vậy $J(t)=2-\frac{1}{t}|(lnt+1)^2+1| (1)$
Từ $(1)$, suy ra $J\left( t \right) < 2$ với $\forall t > 1$
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Trả lời 13-07-12 12:43 PM
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